The greatest height to which a boy can throw a stone is (h). What will be the greatest distance on horizontal surface upto which the boy can throw the stone with the same speed ? Neglect the air friction.

Options

A. `(h)/(2)`
B. h
C. 2h
D. 3h

Explanation

To solve the problem, we need to find the greatest distance (range) on a horizontal surface to which a boy can throw a stone, given that the maximum height he can throw it to is \( h \). We will use the principles of projectile motion.

### Step-by-Step Solution:

1. **Understand the relationship between height and initial velocity:**
The maximum height \( h \) reached by a projectile is given by the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of projection.

2. **Find the maximum height when the angle is 90 degrees:**
The maximum height is achieved when the angle \( \theta = 90^\circ \). Thus, we have:
\[ h = \frac{u^2 \sin^2 90^\circ}{2g} \]
Since \( \sin 90^\circ = 1 \), this simplifies to:
\[ h = \frac{u^2}{2g} \]

3. **Relate initial velocity to height:**
From the equation above, we can express \( u^2 \) in terms of \( h \):
\[ u^2 = 2gh \]

4. **Find the range of the projectile:**
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
For maximum range, the angle \( \theta \) should be \( 45^\circ \):
\[ R = \frac{u^2 \sin 90^\circ}{g} = \frac{u^2}{g} \]

5. **Substitute the value of \( u^2 \):**
Now, substitute \( u^2 = 2gh \) into the range formula:
\[ R = \frac{2gh}{g} = 2h \]

6. **Conclusion:**
Therefore, the greatest distance on the horizontal surface to which the boy can throw the stone with the same speed is:
\[ R = 2h \]

### Final Answer:
The greatest distance on the horizontal surface is \( 2h \).

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